Chapter 4B. Friction and Equilibrium. A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University 2007

Transcription

1 Chapter 4B. Frictio ad Equilibrium A oweroit resetatio by aul E. Tippes, rofessor of hysics Souther olytechic State Uiversity 2007

2 Equilibrium: Util motio begis, all forces o the mower are balaced. Frictio i wheel bearigs ad o the groud oppose the lateral motio.

3 Objectives: After completig this module, you should be able to: Defie ad calculate the coefficiets of kietic ad static frictio, ad give the relatioship of frictio to the ormal force. Apply the cocepts of static ad kietic frictio to problems ivolvig costat motio or impedig motio.

4 Frictio Forces Whe two surfaces are i cotact, frictio forces oppose relative motio or impedig motio. Frictio forces are parallel to the surfaces i cotact ad oppose motio or impedig motio. Static Frictio: No relative motio. Kietic Frictio: Relative motio.

5 Frictio ad the Normal Force 4 N 2 N 8 N 4 N 12 N 6 N The force required to overcome static or kietic frictio is proportioal to the ormal force,. ff s s = s s ff k k = k k

6 Frictio forces are idepedet of area. 4 N 4 N If the total mass pulled is costat, the same force (4 N) is required to overcome frictio eve with twice the area of cotact. For this to be true, it is essetial that ALL other variables be rigidly cotrolled.

7 Frictio forces are idepedet of temperature, provided o chemical or structural variatios occur. 4 N 4 N Heat ca sometimes cause surfaces to become deformed or sticky. I such cases, temperature ca be a factor.

8 Frictio forces are idepedet of speed. 5 m/s 20 m/s 2 N 2 N The force of kietic frictio is the same at 5 m/s as it is for 20 m/s.. Agai, we must assume that there are o chemical or mechaical chages due to speed.

9 The Static Frictio Force Whe a attempt is made to move a object o a surface, static frictio slowly icreases to a MAXIMUM value. f s W I this module, whe we use the followig equatio, we refer oly to the maximum value of static frictio ad simply write: f s s ff s s = s s

10 Costat or Impedig Motio For motio that is impedig ad for motio at costat speed, the resultat force is zero ad F F = 0. (Equilibrium) f s f k Rest f s = 0 Costat Speed f k = 0 Here the weight ad ormal forces are balaced ad do ot affect motio.

11 Frictio ad Acceleratio Whe is greater tha the maximum f s the resultat force produces acceleratio. f k a Costat Speed This case will be discussed i a later chapter. f k = k Note that the kietic frictio force remais costat eve as the velocity icreases.

12 EXAMLE 1: If k = 0.3 ad s = 0.5, what horizotal pull is required to just start a 250-N block movig? f s W + 1. Draw sketch ad free- body diagram as show. 2. List gives ad label what is to be foud: k = 0.3; s = 0.5; W = 250 N Fid: =? to just start 3. Recogize for impedig motio: f s = 0

13 EXAMLE 1(Cot.): s = 0.5, W = 250 N. Fid to overcome f s (max).. Static frictio applies. f s + For this case: f s = 0 4. To fid we eed to kow f s, which is: 250 N f s = =? s 5. To fid : F y = 0 W = 0 W = 250 N 250 N = 250 N (Cotiued)

14 EXAMLE 1(Cot.): s = 0.5, W = 250 N. Fid to overcome f s (max).. Now we kow = 250 N. N 6. Next we fid f s from: f s = s = 0.5 (250 N) 7. For this case: f s = 0 = f s = 0.5 (250 N) = 125 N f s 250 N s = This force (125( N) is eeded to just start motio. Next we cosider eeded for costat speed.

15 EXAMLE 1(Cot.): If k = 0.3 ad s = 0.5, what horizotal pull is required to move with costat speed?? (Overcomig kietic frictio) F y = mam y = 0 k = 0.3 f k mg + - W = 0 = W Now: f k = k = k W F x = 0; - f k = 0 = f k = k W = (0.3)(250 N) = 75.0 N

16 The Normal Force ad Weight The ormal force is NOT always equal to the weight. The followig are examples: W m 30 0 Here the ormal force is less tha weight due to upward compoet of. W Here the ormal force is equal to oly the compoet of weight perpedicular to the plae.

17 Review of Free-body Diagrams: For Frictio roblems: Read problem; draw ad label sketch. Costruct force diagram for each object, vectors at origi of x,y axes. Choose x or y axis alog motio or impedig motio. Dot i i rectagles ad label x ad y compo- ets opposite ad adjacet to agles. Label all compoets; choose positive directio.

18 For Frictio i Equilibrium: Read, draw ad label problem. Draw free-body diagram for for each body. Choose x or or y-axis y alog motio or or impedig motio ad choose directio of of motio as as positive. Idetify the ormal force ad write oe of of followig: f s = s or f k = k f s = s or f k = k For equilibrium, we write for for each axis: F F x = x 0 F F y = y 0 Solve for for ukow quatities.

19 Example 2. A force of 60 N drags a 300-N block by a rope at a agle of 40 0 above the horizotal surface. If u k = 0.2, what force will produce costat speed? W = 300 N f k m W =? 40 0 The force is to be replaced by its com- poets x ad y. 1. Draw ad label a sketch of the problem. 2. Draw free-body diagram. si 40 0 y 40 0 x y f k cos 40 0 W +

20 Example 2 (Cot.). =?; W = 300 N; u k = Fid compoets of : x = cos 40 0 = y = si 40 0 = x = 0.766; y = si f k mg + cos 40 0 Note: Vertical forces are balaced, ad for costat speed, horizotal forces are balaced. Fx 0 Fy 0

21 Example 2 (Cot.). =?; W = 300 N; u k = 0.2. x = y = Apply Equilibrium co- ditios to vertical axis. F y y = N 40 0 f k N N= 0 [ y ad are up (+)]( = 300 N 0.643; Solve for i terms of = 300 N 0.643

22 Example 2 (Cot.). =?; W = 300 N; u k = 0.2. = 300 N Apply F x = 0 to co- stat horizotal motio. F x x = ff k k = N 40 0 f k f k = k = (0.2)(300 N ) f k = (0.2)(300 N ) ) = 60 N f k = 0; (60 N 0.129) = 0

23 Example 2 (Cot.). =?; W = 300 N; u k = N 40 0 f k (60 N )=0 6. Solve for ukow N = = 60 N If = 67 N, the block will be = 60 N dragged at a = 60 N costat speed. = 67.0 N = 67.0 N

24 Example 3: What push up the iclie is eeded to move a 230-N block up the iclie at costat speed if k = 0.3? Step 1: Draw free-body icludig forces, agles ad compoets. y W si 60 0 f k 60 0 x W cos W =230 N Step 2: F y = 0 W cos 60 0 = 0 = (230 N) cos N = 115 N

25 Example 3 (Cot.): Fid Fid to give move up the iclie (W = 230 N). y W si 60 0 f k W x 60 0 W cos 60 0 = 115 N 60 0 W = 230 N Step 3. Apply F x = 0 - f k - W si 60 0 = 0 f k = k = 0.2(115 N) f k = 23 N, =? - 23 N - (230 N)si 60 0 = 0-23 N N N= 0 = 222 N

26 Summary: Importat oits to Cosider Whe Solvig Frictio roblems. The maximum force of static frictio is the force required to just start motio. f s W f s s Equilibrium exists at that istat: F 0; F 0 x y

27 Summary: Importat oits (Cot.) The force of kietic frictio is that force required to maitai costat motio. f k W f k k Equilibrium exists if speed is costat, but f k does ot get larger as the speed is icreased. F 0; F 0 x y

28 Summary: Importat oits (Cot.) Choose a x or y-axis alog the directio of motio or impedig motio. k = 0.3 f k W + The F will be zero alog the x-axis ad alog the y-axis. I this figure, we have: F 0; F 0 x y

29 Summary: Importat oits (Cot.) Remember the ormal force is ot always equal to the weight of a object. It is ecessary to draw 30 0 m the free-body diagram W ad sum forces to solve for the correct value. F 0; F 0 W x y

30 Summary Static Frictio: No relative motio. ff ss s s Kietic Frictio: Relative motio. ff k k = k k rocedure for solutio of equilibrium problems is the same for each case: F 0 F 0 x y

31 CONCLUSION: Chapter 4B Frictio ad Equilibrium